CPhill

7  =    

10  -  √  [  10  -  10/10 ]  =

10  -  √ [ 10  -  1  ]  =

10  - √9  =

10  -  3  =

7

2)   

The   simplifies to

[ bc + 1 ]  /  [ abc + a + c ]  =  3 / 8      which implies that

[ bc + 1 ]  =  3       and

[ a (bc + 1)  + c  ]  =  8

If  a,b,c  do not have to hold unique values....one solution  is  

a  = 2   b  = 1   and c  = 2

So.....  abc  =   2 * 1 * 2   =    4

Here's another way

.726...... [ 6 repeating  ]

Forget the decimal

Subtract  the   non-repeating part from the whole  = 726 - 72  = 654    (1)

Write a number that  has the same number of 9's  as the number of digits in the repeating part and append to this  the same number  of 0's as the number of digits in the non-repeating part  = 900   (2)

Put   (1)  over (2)   =   654/900  = 109/150

The length  is   just the quarter perimeter of a circle with a radius of  8 ft  =

S  =  r θ

 (8) * pi / 2    =

4 pi    ft

(a^2 + 2a)x^2 + (3a)x+1 = 0

This will  have no real solutions  when

(3a)^2 -  4 (a^2 + 2a)*1  < 0     simplify

9a^2  -  4a^2 - 8a   <  0

5a^2  -  8a  <  0

a (5a  -  8 )  <  0     (1)

Let us solve this

a (5a  -  8)  =  0

Setting each factor  to 0   and solve for a

a  = 0        a  =  8/5 

We  have three  possible solution intervals

(-inf, 0)  ,  (0, 8/5)  and  (8/5, inf )     

Note that the middle interval  is the only one that makes  (1)  true

So......the solution  is    0 ≤ a ≤ 8/5 

To solve this, set the functions equal

2x^2 -  10x - 10  =  x^2  - 4x  +  6       rearrange as

x^2 -  6x -  16  =  0      factor as

(x - 8)  ( x + 2)  = 0    

Setting each  factorto 0 and solving for x, we have that x  = 8  and x  = -2

So....using   x^2  - 4x +  6   we can find the two y values

(-2)^2  -4(-2)  +  6  =  18  ⇒  (-2, 18)

8^2  -  4(8)   +  6    =  38  ⇒  (8, 38)

y^4 - 3y^2 - 270   =  0      factor as

(y^2 - 18)  (y^2 + 15)  =  0

Only the first factor will result in a real (positive) answer

So

x^2  - 18  = 0

x^2  =  18          take the positive root

x  = √18    

√3  =  x / (3/√2)

And because we have an equilateral triangle we can find  (1/2) base ,x,  thusly

tan 60  = √18/ x   ⇒   x  =  √18/tan (60)  

So.....the area  is  

(1/2)base  * (height)   =

√18 * √18  /  tan (60)  =

18    / √3   =  

18√3/ 3  =

6√3  units^2

My  answer was the same as EPs   =  9.8  +  l - 5.8 l   =  15.6

His solution is probably  more elegant, though  !!!

\($x^2+y^2=8x-6y-20$\)

Write as

x^2  -  8x  +  y^2   +  6y   =  -20            

Complete the square on x and y

x^2  - 8x  + 16  +  y^2  + 6y  +  9   =  -20  + 16 + 9

(x - 4)^2   +  (y  + 3)^2  =  5

The center is   (4, -3)  =  (x, y)

So   

x + y  =   4  -  3     =   1

2 )

How many three-digit palindromes (numbers that read the same forward and backward) satisfy the following property: the sum of three things -- the hundreds digit, the units digit, and the product of the units and tens digits -- is eight more than the tens digit?

Let  a =  100's digit      b  =  10's digit       c  = units digit

And we have that

a + c + bc   =  b + 8             

a + c +  bc - b =     8

a + c  +  b (c - 1)   = 8

Here  are the possibilities

a    b     c

2    4     2

3    1     3

4    0     4